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我編寫了一個代碼，讓用戶爲兩個選項中的一個投票，並使用修改後的Elo評級系統計算得分。但是，當我爲任一選項投票時，小數點右側的數字將被忽略。我添加了floatval函數，但它沒有幫助。PHP/MySQL忽略小數點右邊的數字
$query = "SELECT pic,score,id FROM nfl
JOIN (SELECT r1.random_id
FROM nfl_map AS r1
JOIN (SELECT (RAND() *
(SELECT MAX(row_id)
FROM nfl_map)) AS row_id)
AS r2
WHERE r1.row_id >= r2.row_id
ORDER BY r1.row_id ASC
LIMIT 1) as rows ON (id = random_id)";
$query_2 = "SELECT pic,score,id FROM nfl
JOIN (SELECT r1.random_id
FROM nfl_map AS r1
JOIN (SELECT (RAND() *
(SELECT MAX(row_id)
FROM nfl_map)) AS row_id)
AS r2
WHERE r1.row_id >= r2.row_id
ORDER BY r1.row_id ASC
LIMIT 1) as rows ON (id = random_id)";
$res_2 = $mysqli->query($query_2);
if (!$res_2) die ("Result display failed: " . $mysqli->errno . " - " . $mysqli->error);
$pic_2 = $res_2->fetch_assoc();
$id_2 = $res_2->fetch_assoc();
$score_2 = $res_2->fetch_assoc();
$res_1 = $mysqli->query($query);
if (!$res_1) die ("Result display failed: " . $mysqli->errno . " - " . $mysqli->error);
$pic_1 = $res_1->fetch_assoc();
$id_1 = $res_1->fetch_assoc();
$score_1 = $res_1->fetch_assoc();
$Ra = $score_1;
$Rb = $score_2;
$calc_pow_1 = (($Rb - $Ra)/400);
$calc_pow_2 = (($Ra - $Rb)/400);
$float_calc_pow_1 = floatval($calc_pow_1);
$float_calc_pow_2 = floatval($calc_pow_2);
$Ea = 1/(1 + pow(10,$float_calc_pow_1));
$Eb = 1/(1 + pow(10,$float_calc_pow_2));
$float_Ea = floatval($Ea);
$float_Eb = floatval($Eb);
// if user votes for picture no. 1
if (isset($_POST['vote_1']) && isset($_POST['id']) && isset($_POST['score']))
{
$id = $_POST['id'];
/* $score = $_POST['score'];
$pic = $_POST['pic']; */
//$mod = 2 * $Eb;
$K = 4;
$float_mod_1 = floatval($K * (1 - $float_Ea));
$query = "UPDATE nfl SET score=?+$float_mod_1 WHERE id=?";
// $score_new = $_POST['score'] + $mod;
$score_new = $_POST['score'] + $float_mod_1;
$stmt = $mysqli->prepare($query);
$stmt->bind_param('di', $_POST['score'], $_POST['id']);
$stmt->execute();
if ($stmt->errno) {
echo "Vote failed: " . $stmt->error();
}
else echo "Vote succeeded! New score of $id_1 is $score_new!
";
$stmt->close();
}
// fetch picture no. 2
//
// query to the database
/* $query_2 = "SELECT pic,score,id FROM nfl
JOIN (SELECT r1.random_id
FROM nfl_map AS r1
JOIN (SELECT (RAND() *
(SELECT MAX(row_id)
FROM nfl_map)) AS row_id)
AS r2
WHERE r1.row_id >= r2.row_id
ORDER BY r1.row_id ASC
LIMIT 1) as rows ON (id = random_id)";
$res_2 = $mysqli->query($query_2);
if (!$res_2) die ("Result display failed: " . $mysqli->errno . " - " . $mysqli->error);
$pic_2 = $res_2->fetch_assoc();
$id_2 = $res_2->fetch_assoc();
$score_2 = $res_2->fetch_assoc(); */
// if user votes for picture no. 2
if (isset($_POST['vote_2']) && isset($_POST['id']) && isset($_POST['score']))
{
$id = $_POST['id'];
/* $score = $_POST['score'];
$pic = $_POST['pic']; */
//$mod = 2 * $Ea;
$K = 4;
$float_mod_2 = floatval($K * (1 - $float_Eb));
$query = "UPDATE nfl SET score=?+$float_mod_2 WHERE id=?";
// $score_new = $_POST['score'] + $mod;
$score_new = $_POST['score'] + $float_mod_2;
/* $query = "UPDATE nfl SET score=?+1 WHERE id=?";
$score_new = $_POST['score'] + $mod; */
$stmt = $mysqli->prepare($query);
$stmt->bind_param('di', $_POST['score'], $_POST['id']);
$stmt->execute();
if ($stmt->errno) {
echo "Vote failed: " . $stmt->error();
}
else echo "Vote succeeded! New score of $id_2 is $score_new!
";
$stmt->close();
}
以下是完整的代碼，如果你需要看到它：http://snipt.org/vDhj2
代碼辦理情況時，用戶投票的第二個選項是幾乎相同的（除了$ EA的變化，以$ Eb等）。 MySQL中'score'行的屬性如下：
float(8,3) not null
謝謝。
編輯：我沒有得到通知了。
看起來我修正了通知。沒有得到它了。感謝你們三個人。但我該如何解決另一個問題？ – Pawel
@Pawel我不完全確定在這種情況下，但你可能想看看[這個問題和我的答案]（http://stackoverflow.com/questions/12155465/pdo-difference-between-binding-params-and -concatenating-string/12155531＃12155531）處理綁定參數被視爲字符串，除非完全指定爲別的東西。這可能是一個內部mysql問題，導致它做一個整數操作，並在小數點後丟失數據（但我再也不確定這個問題） – Fluffeh
我很新，PHP＆ MySQL的，所以我可能是錯的，但我的代碼明確指出：'$ stmt-> bind_param（'ii'，$ _POST ['score']，$ _POST ['id']）;' 如果這是問題，這一點的代碼修復它？ – Pawel